7
$\begingroup$

Does a monotonic transformation of a homothetic utility function imply the preference relation on the set of consumption bundles is still homothetic?

Obviously, if a utility function on a set of consumption bundles is homothetic, then the preference relation is homothetic.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
8
$\begingroup$

Yes. We know that a monotonic transformation of a utility function still represents the same preferences and as the old utility function represented homothetic preferences the new one does, too.

As an easy example you could look at Cobb-Douglas utility functions of the form $u(x,y) = a\left(x y\right)^\alpha$. For $\alpha = \frac12$ the utility function is homogeneous of degree 1, but for every $\alpha,a>0,$ the preference relation is homogeneous of degree 1.

We never used homothetic as a property of the utility function to answer your question (as it works for every utility function). A homothetic utility function is, to the best of my knowledge, not very clearly defined, I have see two different versions:

  1. Homothetic as different name for homogeneous of degree 1
  2. A homothetic utility function is a utility function that represents a homothetic preference relation.

Which means, with definition 2 Cobb-Douglas utility functions with equal weights are always homothetic, with definition 2 only some of them are: $$u(tx,ty) = (tx)^\frac12 (ty)^\frac12 = t xy = t u(x,y)$$ and some are not $$u(tx,ty) = (tx) (ty) = t^2 xy \neq t u(x,y)$$

Improve this answer
$\endgroup$
6
  • $\begingroup$ Wait, I thought Cobb-Douglas was homothetic for all the utility functions because $$u(tx,ty) = (tx)^\alpha (ty)^{1-\alpha}= t x^\alpha y^{1-\alpha} $$ is that wrong? Is only the preference relation homothetic for all $\alpha$? If so, why? $\endgroup$
    – Stan Shunpike
    Apr 5, 2015 at 14:14
  • $\begingroup$ @StanShunpike Your CD function and mine are not the same, they are both special cases. In mine both exponents are the same, in yours they add up to 1. I actually don't use homothetic for utility functions, as people seem to use different definitions for homothetic utility functions: One being "its preference relation is homothetic" the other one is "it is homogeneous of degree 1", I think I will try to make that clear in the answer. Just do the same calculations you have done with your utility function with mine (and $\alpha =1$), then you will see it is not homogeneous of degree 1. $\endgroup$
    – The Almighty Bob
    Apr 5, 2015 at 14:58
  • $\begingroup$ So, I hope this helps. $\endgroup$
    – The Almighty Bob
    Apr 5, 2015 at 15:10
  • $\begingroup$ So if I understand you correctly, in your answer you showed the special cases you and I mentioned are homothetic utility functions. They therefore have a homothetic preference relation. Is that the basic idea? $\endgroup$
    – Stan Shunpike
    Apr 5, 2015 at 17:28
  • $\begingroup$ Mine is only homogeneous of degree 1 for the case $\alpha = 1/2$, yours always is. But having one is enough, as every other of mine is just a monotonic transformation of the one that is. But yes, it is the basic idea. $\endgroup$
    – The Almighty Bob
    Apr 5, 2015 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.